# Fair Coin Toss?

Discussion in 'Probability' started by nstyrthy, Sep 15, 2008.

1. ### nstyrthyGuest

Assuming we flip a fair coin 9 times and get 9 heads. What's the
probability that the coin will come up heads on the next toss?

I'm a li'l confused here. They say a fair coin toss is independent of
other toss, and so the probability should be 50/50. But per Bayes
theorem, we need to revise the probability and that means there's 9/10
chance of getting heads on the next flip (assuming my calculation is
right). But then again, statistics show that of 10 flips, roughly 5
should be heads, and that means, this fair coin is *overdue* for
showing tails, and so has a high probability for coming up tails.

Clearly, I'm wrong somewhere. What am I missing?

Any help will be greatly appreciated. Thanks.

nstyrthy, Sep 15, 2008

2. ### se16Guest

You have said it is a fair coin. So by definition the chance should
be 50% every time.

Your Bayesian analysis would produce the same result, provided that
before the execise started you were absolutely certain that you had a
fair coin (known as a point prior). If however you had been prepared
to entertain some doubt then the chance that next flip would be heads
would be more than 50%. How much more would depend on the nature and
extent of your doubt: for example, if you had thought that there had
been a 0.5% chance of a double-headed coin, a 0.5% chance of a double
tailed coin and a 99% chance of a fair coin then after nine heads you
would revise these to about 72%, 28% and 0% repectively, so your new
belief would be that the next coin had about an 86% chance of being

Your "overdue" thinking contradicts the memoryless nature of coin
flips. You could use such an argument if for example you were taking
balls out of a bag and not putting them back, where you knew that
originally there were equal numbers of red and yellow balls (and you
could calculate a probability if you knew how many there were).

se16, Sep 15, 2008

3. ### illywhackerGuest

Hi. Sorry: I do not know what to call you as you have not signed your
post.

You are indeed confusing three different things.

1) The knowledge you have (i.e. the probability you assign) *before*
you
see the coin tossed, assuming have good reason to believe that the
coin
tossing method is 'fair'. This is the situation in which you assign
probability p = 0.5 to getting heads on each coin toss and treat the
tosses
as independent. In this case:

- The best prediction for the number of heads you will get in ten
tosses is
five.

- The outcomes of the tosses are independent because knowing the
outcome of
one toss will not affect the probability you assign to the outcome of
another, since you assume you know that p = 0.5.

2) The knowledge you have (i.e. the probability you assign) *before*
you
see the coin tossed, assuming now that you have *no* particular
knowledge
about the coin or the coin tossing method, i.e. you do not know if it
is
'fair' or not. In this situation, you do not know what probability to
assign to each coin toss, so you have to sum equally over all
possibilities
for p. In this case:

- The best prediction for the number of heads you will get in ten
tosses is
still five.

- But the outcomes of the tosses are not independent because knowing
the
outcome of one toss would help you to estimate the probability p of
getting
heads on another toss and hence influence your prediction for the
outcome.

3) The knowledge you have (i.e. the probability you assign) *after*
you
have seen the coin come up heads ten times. What you know now
obviously
depends on what you knew before you saw the tosses, as another poster
has
indicated. If you were previously very certain that p = 0.5, i.e. case
(1),
then this might be revised a little, but not much. You will put down
the
ten heads to a 'fluke'. However, if you were previously completely
ignorant
about p, as in case (2), your uncertainty about p has now been
reduced, and
what is more you have good reason to doubt that the coin tossing
method is
'fair'.

Thus this case is intermediate between (1) and (2). In (1), you assume
you
are certain about p. In (2), you assume you know nothing about p, i.e.
you
are maximally uncertain. In (3), even if you start off knowing nothing
about p, the results of the tosses tell you something, although even
so you
cannot be certain. Hence:

- Your best prediction of the number of heads you will get in the next
ten
tosses is no longer five, since the results indicate that p is very
unlikely to be 0.5.

- The residual uncertainty means that seeing the results of more
tosses can
still tell you more about p, so the tosses are still not completely
independent, although more so than in (2).

-------------------------------

The *most important point* here is this: these are *not* statements
the coin, the coin tossing method, physics, or any fact to do with the
world. They are statements about your *knowledge* of these things. If
you
do not get to grips with this idea, you will be coming back here again
and
again asking the same questions that many others have asked, confused
by
their belief that probability is a property of the world, like weight,
rather than a description of our knowledge of the world.

Luckily this idea is very intuitive, so intuitive in fact that people
taking statistics courses have to have it beaten out of them by
intellectual intimidation, and even then can barely stop themselves
from
using the forms of language to which it gives rise, e.g. 'the
probability
of the parameter value'. See any of the many discussions concerning
'how to
talk about confidence intervals' to get an idea of the tortuous nature
of
the alternatives.

illywhacker;

illywhacker, Sep 15, 2008
4. ### nstyrthyGuest

Thanks everyone for the help.

illywhacker, thanks for the very clear and lucid explanation. T

nstyrthy, Sep 16, 2008